The current in the primary circuit of a potentiometer is $0.2 \, A$. The specific resistance and cross-section of the potentiometer wire are $4 \times 10^{-7} \, \Omega \cdot m$ and $8 \times 10^{-7} \, m^2$ respectively. The potential gradient will be equal to .............. $V/m$.

$A$ potentiometer wire has length $4\, m$ and resistance $8\, \Omega$. The resistance that must be connected in series with the wire and an accumulator of e.m.f. $2\, V$,so as to get a potential gradient of $1\, mV$ per $cm$ on the wire is ............. $\Omega$.

$A$ battery of internal resistance $1 \, \Omega$ and $emf$ $3 \, V$ sends a current through $1 \, m$ of uniform wire of resistance $5 \, \Omega$. The poles of a cell of $emf$ $1.4 \, V$ are connected to two points on the wire such that no current passes through this cell. The potential gradient of the wire is:

$A$ Daniel cell is balanced on $125\,cm$ length of a potentiometer wire. Now the cell is short-circuited by a resistance $2\,\Omega$ and the balance is obtained at $100\,cm$. The internal resistance of the Daniel cell is .............. $\Omega$.

$A$ thermocouple of negligible resistance produces an $e.m.f.$ of $40\,\mu V/^{\circ}C$ in the linear range of temperature. $A$ galvanometer of resistance $10\,\Omega$ whose sensitivity is $1\,\mu A/\text{div}$ is employed with the thermocouple. The smallest value of temperature difference that can be detected by the system will be (in $^{\circ}C$)

To determine the internal resistance of a cell by using a potentiometer, the null point is at $1 \, m$ when the cell is shunted by $3 \, \Omega$ resistance and at a length $1.5 \, m$ when the cell is shunted by $6 \, \Omega$ resistance. The internal resistance of the cell is: